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2 years ago

8

A diver runs horizontally with a speed of 1.20 m/s off a platform that is 10.0 m above the water. What is his speed just before

striking the water

1 answer:

DENIUS [597]2 years ago

4 0

Answer:

14.05m/s

Explanation:

We are given that

Horizontal speed of diver= $v_x=1.2m/s$

Distance, y=-10 m

It is taken as negative because the diver goes downward.

There is no air resistance therefore, there is acceleration due to gravity .

We know that

$g=-9.8 m/s^2$

Initial vertical velocity, $u_y=0$

$v^2-u^2=2as$

Using the formula

$v^2_y-0=2\times (-9.8)(-10)$

$v_y=\sqrt{2(-9.8)(-10)}$

$v_y=14 m/s$

$v=\sqrt{v^2_x+v^2_y}$

Using the formula

$v=\sqrt{(1.2)^2+(14)^2}=14.05m/s$

Hence, the speed of diver before just striking the water=14.05m/s

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At time t, gives the position of a 3.0 kg particle relative to the origin of an xy coordinate system ( ModifyingAbove r With rig

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Complete Question

The complete Question is shown on the first uploaded image

Answer:

a

The torque acting on the particle is $\tau = 48t \r k$

b

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Explanation:

From the equation we are told that

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The torque acting on the particle is mathematically represented as

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$\r v = \frac{d \r r }{dt}$

Substituting for $\r r$

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Now the cross product of $\r r \ and \ \r v$ is mathematically evaluated as

$\r r \ \ X \ \ \r v = \left[\begin{array}{ccc}{\r i}&{\r j}&{\r k}\\{4t^2}&{-2t -6t^2}&0\\{8t}&{-2 -12t}&0\end{array}\right]$

$= 0 \r i + 0 \r j + (- 8t^2 -48t^3 + 16t^2 + 48t^3 ) \r k$

$\r r \ \ X \ \ \r v = 8t^2 \r k$

So the angular momentum becomes

$\r l = m (8t^2 \r k)$

Substituting for m

$\r l = 3 * (8t^2 \r k)$

$\r l =24t^2 \r k$

Substituting into equation for torque

$\tau = \frac{d}{dt} [24t^2 \r k]$

$\tau = 48t \r k$

The magnitude of the angular momentum can be evaluated mathematically as

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1 year ago

A toy rocket launcher can project a toy rocket at a speed as high as 35.0 m/s.

Anestetic [448]

Answer:

(a) 62.5 m

(b) 7.14 s

Explanation:

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Use third equation of motion

$v^{2}=u^{2}+2as$

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h = 62.5 m

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